103 Binary Tree Zigzag Level Order Traversal

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        vector<vector<int> > result;
        if (root == NULL) return result;
        
        bool rev = false;
        queue<TreeNode*> q;
        q.push(root);
        
        while (!q.empty()) {
            int count = q.size();
            vector<int> ans;
            for (int i = 0; i < count; i++) {
                TreeNode *tmp = q.front(); q.pop();
                ans.push_back(tmp->val);
                if (tmp->left != NULL) q.push(tmp->left);
                if (tmp->right != NULL) q.push(tmp->right);
            }
            
            if (rev) reverse(ans.begin(), ans.end());
            rev = !rev;
            result.push_back(ans);
        }
        
        return result;        
    }
};